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kjetil b halvorsen ♦
kjetil b halvorsen ♦
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The point is that, when one says that "The poisson distribution do not have constant variance" (for instance), one is referring to the POisson distribution family. In a statistical analysis, say, we assume that the count we are interested in do have some poisson distribution, with expectation $\lambda$, for some value of $\lambda >0$, but we do not specify the value of $\lambda$, we do estimate $\lambda$ from the data. Since the variance is then also $\lambda$, it is not constant as a function of the mean. The same goes for, mutatis mutandi, the binomial or bernoulli distribution.

Comparing with the situation for normal distributed models, there we have two parameters, $\mu, \sigma^2$, modelling expectation and variance separately. So then, it is a mathematical possibility that the expectation can vary, without the variance varying with it. So, the variance, as a function of the mean, can be a constant. That is mathematically impossible for a poisson model.

So if we, say, want to compare two treatmente (or treatment with a control), if the response variable is poisson distributed, it is mathematically impossible for the treatment to change the mean, without it at the same time changeing the variance. If the response in noramlly distributed, it is mathematically possible for the treatment to change the mean, without it also changing the variance (of course, in a specific experiment, it might well be that the treatment also change the variance).

EDIT

The OP asks in a comment: " "one is referring to the Poisson distribution family" I assume this means, that we have not fixed λ to a specific value but are speaking about all possible infinite λ, and if we fix a value then we have a single variance. Can you confirm? ". Yes, of course, if you specify a fixed value for $\lambda$, say as an example $\lambda=10$, then the variance is constant! The non-constancy comes from not knowing $\lambda$. But, if you specify $\lambda$, then you know completely the distribution of your data (or, at least, you are saying that you know it ...), and then there is no statistics problem left! Then you have a pure probablity problem, not a statistics problem.

The point is that, when one says that "The poisson distribution do not have constant variance" (for instance), one is referring to the POisson distribution family. In a statistical analysis, say, we assume that the count we are interested in do have some poisson distribution, with expectation $\lambda$, for some value of $\lambda >0$, but we do not specify the value of $\lambda$, we do estimate $\lambda$ from the data. Since the variance is then also $\lambda$, it is not constant as a function of the mean. The same goes for, mutatis mutandi, the binomial or bernoulli distribution.

Comparing with the situation for normal distributed models, there we have two parameters, $\mu, \sigma^2$, modelling expectation and variance separately. So then, it is a mathematical possibility that the expectation can vary, without the variance varying with it. So, the variance, as a function of the mean, can be a constant. That is mathematically impossible for a poisson model.

So if we, say, want to compare two treatmente (or treatment with a control), if the response variable is poisson distributed, it is mathematically impossible for the treatment to change the mean, without it at the same time changeing the variance. If the response in noramlly distributed, it is mathematically possible for the treatment to change the mean, without it also changing the variance (of course, in a specific experiment, it might well be that the treatment also change the variance).

The point is that, when one says that "The poisson distribution do not have constant variance" (for instance), one is referring to the POisson distribution family. In a statistical analysis, say, we assume that the count we are interested in do have some poisson distribution, with expectation $\lambda$, for some value of $\lambda >0$, but we do not specify the value of $\lambda$, we do estimate $\lambda$ from the data. Since the variance is then also $\lambda$, it is not constant as a function of the mean. The same goes for, mutatis mutandi, the binomial or bernoulli distribution.

Comparing with the situation for normal distributed models, there we have two parameters, $\mu, \sigma^2$, modelling expectation and variance separately. So then, it is a mathematical possibility that the expectation can vary, without the variance varying with it. So, the variance, as a function of the mean, can be a constant. That is mathematically impossible for a poisson model.

So if we, say, want to compare two treatmente (or treatment with a control), if the response variable is poisson distributed, it is mathematically impossible for the treatment to change the mean, without it at the same time changeing the variance. If the response in noramlly distributed, it is mathematically possible for the treatment to change the mean, without it also changing the variance (of course, in a specific experiment, it might well be that the treatment also change the variance).

EDIT

The OP asks in a comment: " "one is referring to the Poisson distribution family" I assume this means, that we have not fixed λ to a specific value but are speaking about all possible infinite λ, and if we fix a value then we have a single variance. Can you confirm? ". Yes, of course, if you specify a fixed value for $\lambda$, say as an example $\lambda=10$, then the variance is constant! The non-constancy comes from not knowing $\lambda$. But, if you specify $\lambda$, then you know completely the distribution of your data (or, at least, you are saying that you know it ...), and then there is no statistics problem left! Then you have a pure probablity problem, not a statistics problem.

Source Link
kjetil b halvorsen ♦
kjetil b halvorsen ♦
  • 82.8k
  • 32
  • 201
  • 663

The point is that, when one says that "The poisson distribution do not have constant variance" (for instance), one is referring to the POisson distribution family. In a statistical analysis, say, we assume that the count we are interested in do have some poisson distribution, with expectation $\lambda$, for some value of $\lambda >0$, but we do not specify the value of $\lambda$, we do estimate $\lambda$ from the data. Since the variance is then also $\lambda$, it is not constant as a function of the mean. The same goes for, mutatis mutandi, the binomial or bernoulli distribution.

Comparing with the situation for normal distributed models, there we have two parameters, $\mu, \sigma^2$, modelling expectation and variance separately. So then, it is a mathematical possibility that the expectation can vary, without the variance varying with it. So, the variance, as a function of the mean, can be a constant. That is mathematically impossible for a poisson model.

So if we, say, want to compare two treatmente (or treatment with a control), if the response variable is poisson distributed, it is mathematically impossible for the treatment to change the mean, without it at the same time changeing the variance. If the response in noramlly distributed, it is mathematically possible for the treatment to change the mean, without it also changing the variance (of course, in a specific experiment, it might well be that the treatment also change the variance).